Talk:Tetration
Here, it uses what looks like circle notation - can circle notation accept parameters? If it can, this should be mentioned on Circle notation. Circle notation, an extension to Steinhaus-Moser Notation, is entirely different from Rob Munafo's hyper-operators. [[User:Followed by 100 zeroes|'Followed by 100 zeroes']] (talk | ) 21:01, 7 March 2009 (UTC) I've been trying for a while to decode Geisler's site, but the math is giving me migraines -.- Is anyone interested in perusing that? FB100Z • talk • 03:09, April 29, 2012 (UTC) Small examples only? Well, I think that only giving small examples doesn't tell you how large the larger numbers are generated by tetration, so it won't be helpful. Large examples tell you how this function is important in googology. Cloudy176 (talk) 10:38, November 18, 2012 (UTC) :Okay, you're right. Some examples reaching into scientific notation and beyond would be nice, but don't let's fill the page with seas of digits. FB100Z • talk • 20:56, November 18, 2012 (UTC) Real tetration It's don't seem to me that \(^{1/x}a\) = \(\sqrtx{a}_s\). Just see that the infinith-super root of 2 = \(\sqrt{2}\), but \(^{0}a\) = 1, and 1/0 = infinity. Ikosarakt1 (talk) 13:11, January 13, 2013 (UTC) What's about defining \(a \uparrow\uparrow (1/b)\) as \(\sqrt\uparrow\uparrow (b-1)(a)\). That's reasonable, since for b=1 it gives a, and as \(b \rightarrow \infty\) it tends to 0. Last digits convergence We know very well that for integers \(n\), the last digits of power towers of \(n\) converge, making it easy to predict the last digits of a very large power tower (such as a chained arrow or BEAF array). For example, a power tower of threes converges to ...04575627262464195387, the final digits of Graham's number. There's a page on the InterNet about "oology," which explores infinities recreationally. (Perhaps coincidentally, the page is from the André Joyce Fan Club, from which the term "googology" first originated.) It has silly things like turning fractions backwards: \(1/3 = 0.333\ldots\), so \(3\backslash 1\) ("3 conquered by 1," as in "divide and conquer") is \(\ldots 333\). Although clearly this is recreational mathematics and somewhat tongue-in-cheek, these look a lot like the notation for digit convergence in tetration. This can give us funny-looking equations such \(^\infty3 = \ldots 04575627262464195387\). FB100Z • talk • 02:08, February 17, 2013 (UTC) :If I take the conquer operator for granted, \(x/1\) is the reversal of \(1\backslash x\). So I guess we have \(1\backslash ^\infty3 = 0.78359146426272657540\ldots\) okay, what the hell did I just write FB100Z • talk • 02:12, February 17, 2013 (UTC) :I've typed a simple-math book about integer tetration, exploring its p-adic convergence properties. There are also pseudo-convergence properties involving the figures next to the rightmost ones. The caos on the left fight against the order on the right of the number and there are some rules to point out this feature. In short, it's easy to predict the digit on the left of the leftmost frozen digit (without calculating tetration or using modular arithmetics and cycles), but it's very complex to predict other digits, even if you have already found their mutual constraints. :SPIqr (talk) 02:14, February 19, 2013 (UTC) In popular culture I found a rare occurrence of tetration humor: http://www.smbc-comics.com/?id=1666 (click on the red button below the comic) FB100Z • talk • 06:04, September 12, 2013 (UTC) :What about Knuth Paper-Stack Notation? -- ☁ I want more ⛅ 07:17, September 12, 2013 (UTC) ::True, true. Still, the SMBC comic may be the world's first sex joke with tetration in it. FB100Z • talk • 07:57, September 12, 2013 (UTC) ::::And then they went on to make one with pentation and hexation in it. ArtismScrub (talk) 02:38, December 1, 2017 (UTC) "natural" tetration is there any tetrational analog to the function \(f(x) = e^x\)? that is, is there a constant \(\nu\) such that \(f(x) = {}^x\nu\) is a "natural" function? FB100Z • talk • 08:58, September 24, 2013 (UTC) :We have to define tetration for real heights before answering this question. Ikosarakt1 (talk ^ ) 15:22, September 24, 2013 (UTC) ::Since e is the infinite sum of the factorial reciprocals, I would say that your \(\nu\) might possibly be the infinite sum of the expofactorial reciprocals. Then again, since ba is not defined for non-integer b, this is meaningless. —Preceding unsigned comment added by ArtismScrub (talk • ) 03:52, January 24, 2018 (UTC) Definition for ordinals I think it's natural to define w^^(1+a) = w^(w^^a), but not w^^(a+1) = w^(w^^a). For finite "a", it doesn't matter, but for a = w there is the fact that w^^(1+w) = w^(w^^w) = w^(e(0)) = e(0) = w^^w. That's consistent with the fact that 1+w = w. For w^^(a+1), we should use Saibian's definition I believe. So w^^(a+1) = {w^^a}^{w}. Don't be confused with something like (w^^a)^w, look how the proper variant works for a = 4: w^^(4+1) = {w^^4}^{w} = {w^w^w^w}^{w} = w^w^w^w^w, but not (w^w^w^w)^w. Ikosarakt1 (talk ^ ) 07:56, July 27, 2014 (UTC) Nontrivial solutions in the natural numbers to a^^b = c^^d? Are there numbers expressible as a tetration in more than one way? I think I have proven that there are none, but I'm not enirely sure. Full proof with all details would be quite lengthly. LittlePeng9 (talk) 15:59, August 6, 2014 (UTC) Could you say what your approach was? Suppose that we had a solution. Then we would have a^(a^^(b-1))=c^(c^^(d-1)). Every solution of p^q=r^s has p and r being integer powers of some number k. So we have a=k^l, c=k^m. From non-triviality k isn't 1. Now by laws of exponentiation we can get l*(k^l)^^(b-2)=m*(k^m)^^(d-2). If we assume l>=m then we get l/m is an integer and a power of k. By further manipulation we get that l must be divisible by m*(k^m)^^(d-2), which leads to the contradiction. I used induction to show this, and I'm not entirely sure if this will work. LittlePeng9 (talk) 18:21, August 6, 2014 (UTC) Actually, it is l*(k^l)^^(b-1) = m*(k^m)^^(d-1). Continuing, let l = m*k^n. Then (m*k^n)((k^(m*k^n))^^(b-1)) = m*(k^m)^^(d-1), and therefore n + (m*k^n)*((k^(m*k^n))^^(b-2)) = m*((k^m)^^(d-2)). Thus n = m*((k^m)^^(d-2)) - (m*k^n)*((k^(m*k^n))^^(b-2)) = m*k^u - m*k^v, where we have v >= n and therefore u >= n. So k^n divides n, which is impossible. This is of course assuming a,b,c,d > 1 and a != c. Deedlit11 (talk) 03:23, August 7, 2014 (UTC) *** I attempted to create a redirect there, but this was blocked by a spam filter. -- 21:36, February 24, 2018 (UTC)